Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an open
is required to reflect power and I'm searching for which it is, an open or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or analogy
and some math wouldn't hurt.

Hank, EM reflections are covered on this web page.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

In particular: "Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists between
reflections of p, then reflected wavefronts interfere destructively, and
overall reflected intensity is a minimum. If the two reflections are of
equal amplitude, then this amplitude (and hence intensity) minimum will
be zero."

"In the absence of absorption or scatter, the principle of conservation of
energy indicates all "lost" reflected intensity will appear as enhanced
intensity in the transmitted beam. The sum of the reflected and transmitted
beam intensities is always equal to the incident intensity. This important
fact has been confirmed experimentally."

In order for (rearward-traveling) "reflected intensity" to "appear as
(forward-traveling) enhanced intensity in the transmitted beam", the
momentum of that (rearward-traveling) intensity must change directions.
Thus, it appears that total destructive interference between two rearward-
traveling reflected waves in a transmission line will reverse the direction
of momentum of the energy in those canceled reflected waves.

We need to change a few of your statements:

Any power not dissipated or radiated by an antenna is reflected back.
"Dissipation" means EM energy transformed into heat, according to
the IEEE Dictionary.

The transmitter/tuner end will not re-reflect 100% of the reflected energy
unless there exists a short, open, pure reactance, or "total destructive
interference" as explained in _Optics_, by Hecht.

Besides a short or an open, a purely reactive impedance will cause
100% energy reflection. Apparently, so will "total destructive
interference". Quoting from _Microwave_Transmission_, by J. C. Slater:

"The method of eliminating reflections is based on the interference
between waves. ... The fundamental principle behind the elimination
of reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

That's pretty clear. We get one set of rearward-traveling reflections
at the match point. We get another set of rearward-traveling reflections
at the antenna. If these two sets of reflections are equal in magnitude
and opposite in phase at the match point, they cancel each other and the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Henry Kolesnik wrote:

I know that any power not dissipated by an antenna is reflected back to
the
transmitter. Then the transmitter "reflects" this reflection back to
antenna, ad nauseum until its all gone. I also know that a short or an
open
is required to reflect power and I'm searching for which it is, an open
or a
short. I'm inclined to think it's a virtual open but I'm at a loss to
understand that and I wonder if someone has a good explanation or
analogy
and some math wouldn't hurt.

Hank, EM reflections are covered on this web page.

http://www.mellesgriot.com/products/optics/oc_2_1.htm

In particular: "Clearly, if the wavelength of the incident light and the
thickness of the film are such that a phase difference exists between
reflections of p, then reflected wavefronts interfere destructively, and
overall reflected intensity is a minimum. If the two reflections are of
equal amplitude, then this amplitude (and hence intensity) minimum will
be zero."

"In the absence of absorption or scatter, the principle of conservation of
energy indicates all "lost" reflected intensity will appear as enhanced
intensity in the transmitted beam. The sum of the reflected and
transmitted
beam intensities is always equal to the incident intensity. This important
fact has been confirmed experimentally."

In order for (rearward-traveling) "reflected intensity" to "appear as
(forward-traveling) enhanced intensity in the transmitted beam", the
momentum of that (rearward-traveling) intensity must change directions.
Thus, it appears that total destructive interference between two rearward-
traveling reflected waves in a transmission line will reverse the
direction
of momentum of the energy in those canceled reflected waves.

We need to change a few of your statements:

Any power not dissipated or radiated by an antenna is reflected back.
"Dissipation" means EM energy transformed into heat, according to
the IEEE Dictionary.

The transmitter/tuner end will not re-reflect 100% of the reflected energy
unless there exists a short, open, pure reactance, or "total destructive
interference" as explained in _Optics_, by Hecht.

Besides a short or an open, a purely reactive impedance will cause
100% energy reflection. Apparently, so will "total destructive
interference". Quoting from _Microwave_Transmission_, by J. C. Slater:

"The method of eliminating reflections is based on the interference
between waves. ... The fundamental principle behind the elimination
of reflections is then to have each reflected wave canceled by another
wave of equal amplitude and opposite phase."

That's pretty clear. We get one set of rearward-traveling reflections
at the match point. We get another set of rearward-traveling reflections
at the antenna. If these two sets of reflections are equal in magnitude
and opposite in phase at the match point, they cancel each other and the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.
--
73, Cecil http://www.qsl.net/w5dxp
Cecil,

I am not quite sure what you are saying. But, I ran a SPICE simulation of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission line
shorted at the far end. Independent of R0, in steady state the voltage at
the input end of the transmission line will be 1V. The effect of R0 is to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.

Tam/WB2TT wrote:
I am not quite sure what you are saying. But, I ran a SPICE simulation of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission line
shorted at the far end. Independent of R0, in steady state the voltage at
the input end of the transmission line will be 1V. The effect of R0 is to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.

Does SPICE report the steady-state forward and reflected waves
or just the superposition of those two waves? We all know what
they look like when superposed. The question is whether the
identity of the forward and reflected waves disappear after
they are superposed. To the best of my knowledge, the very
existence of standing waves requires the existence of a forward-
traveling wave and a rearward-traveling wave.

I have asked for examples of standing waves void of rearward-
traveling waves and none has been forthcoming.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Tam/WB2TT wrote:
I am not quite sure what you are saying. But, I ran a SPICE simulation
of
the following:
1V 1MHz source with resistor R0 feeding a 50 Ohm 250 ns transmission
line
shorted at the far end. Independent of R0, in steady state the voltage
at
the input end of the transmission line will be 1V. The effect of R0 is
to
limit how long it takes to reach steady state. For R0 = 50 Ohms, it is
one
cycle; for R0 = 500 Ohms, it is about 8 cycles, as eyeballed off the
waveform display.

Does SPICE report the steady-state forward and reflected waves
or just the superposition of those two waves? We all know what
they look like when superposed. The question is whether the
identity of the forward and reflected waves disappear after
they are superposed. To the best of my knowledge, the very
existence of standing waves requires the existence of a forward-
traveling wave and a rearward-traveling wave.

I have asked for examples of standing waves void of rearward-
traveling waves and none has been forthcoming.
--
73, Cecil http://www.qsl.net/w5dxp
It shows the composite voltage waveform, and the net current. Exactly what a
Bird wattmeter would do. Of course the Bird only shows you steady state,
Spice (SWCAD) swhows how you got there.

Tam/WB2TT

Tam/WB2TT wrote:
It shows the composite voltage waveform, and the net current. Exactly what a
Bird wattmeter would do.

That's not what a Bird wattmeter does. A Bird wattmeter possesses
a directional coupler. SPICE apparently does not. Is it possible
to add a directional coupler to SPICE? If you know the Z0, the net
voltage/current magnitudes/phases, it should be possible to use
phasor addition/subtraction to obtain the forward and reflected
components, just like the Bird wattmeter does.
--
73, Cecil http://www.qsl.net/w5dxp



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"Cecil Moore" wrote in message
...
Tam/WB2TT wrote:
It shows the composite voltage waveform, and the net current. Exactly
what a
Bird wattmeter would do.

That's not what a Bird wattmeter does. A Bird wattmeter possesses
a directional coupler.

Sort of. I have built circuits that subtract out the transmitted signal,
leaving the received signal.signal. The Bird is cruder than that.

SPICE apparently does not.

I was driving it with a sine wave, but did a transient analysis. The whole
point is it does not have to know about reflections. It calculates the
waveform by using the same rules that are used to derive standing waves and
reflections.

Is it possible
to add a directional coupler to SPICE?

I have built a SPICE model of a Kenwood power/SWR meter (Have better
schematic than for a Bird). Actually, an idealized version that is not
physically realizable; I did this on purpose. Clearly shows what the
limitations are. Interesting thing is that there is information present that
no SWR meter that I know of displays. For an SWR other than 1:1, you can
deduce whether RL is bigger or smaller than Z0 by comparing two voltages.


If you know the Z0, the net
voltage/current magnitudes/phases, it should be possible to use
phasor addition/subtraction to obtain the forward and reflected
components, just like the Bird wattmeter does.

But there is only one voltage sample, which is the sum of Vf and Vr. There
*are* two current samples, but they are exactly the same, only one is 180
degrees out of phase due to looking at the opposite end of the current
transformer.

Here is what happens. Say you want a meter that shows 100W full scale when
feeding a 50 Ohm load. That is 70.7 V and 1.414 A. In the "Forward"
direction this leads to

100 = 70.7K1 + 1.414K2

In the "Reverse"direction, we know that Pr=0, so

0 = 70.7K1 - 1.414K2 ( The minus sign comes from reversing the current
reading).

You have 2 equations, so you can solve for K1 and K2. You know it can't
*really* measure power, because there is no multiplier. Just like the Bird,
it *adds* (vector wise) voltage and current.
--
73, Cecil http://www.qsl.net/w5dxp



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Tam/WB2TT wrote:
You have 2 equations, so you can solve for K1 and K2. You know it can't
*really* measure power, because there is no multiplier. Just like the Bird,
it *adds* (vector wise) voltage and current.

Exactly, and if you work out the math, you will find it yields
a meter deflection that can be calibrated in watts of forward
or reflected power.

In your example, assume that 70.7v yields a 5v sample and 1.414a
yields a 5v sample. If they are in phase, 10v will indicate 100
watts forward and zero volts will indicate zero watts reflected.
If they are not equal and not in phase, their sum still indicates
forward watts and their difference still indicates reflected watts.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:
the
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.

Yeeesh. You had it on, dog, up until that. And don't try to tell me
(again) that I'm lying that you said it.

(Remember when you wrote this? "If reflected energy makes its way back
into the final amp, it was never generated in the first place, by
definition." Hint: apply the same idea to your "rearward-traveling
momentum" and you'll have it.)

73, Jim AC6XG

Jim Kelley wrote:

Cecil Moore wrote:
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.

Yeeesh. You had it on, dog, up until that. And don't try to tell me
(again) that I'm lying that you said it.

(Remember when you wrote this? "If reflected energy makes its way back
into the final amp, it was never generated in the first place, by
definition." Hint: apply the same idea to your "rearward-traveling
momentum" and you'll have it.)

Egads Jim, exactly how much of reality do you think I am capable of
ignoring? Obviously, not as much as you. Why not just say, "God is
the cause of everything I (Jim) cannot explain or understand."?
The meaning would be virtually identical to your present positions.

I *don't* agree with that definition above and your implication that
momentum and energy don't need to be conserved is simply metaphysics
in action. You can argue against energy conservation all you want.
*Conservation of Momentum* has got you over a barrel in this argument
whether you realize it or not. Somehow, the momentum in the wave
reflected from a mismatched load is reversed. Please explain how
that happens without changing directions. Somehow, the energy in
the wave reflected from a mismatched load changes directions. Please
enlighten us on exactly the mechanism involved. Hint: J.C.Slater
explained it all in _Microwave_Transmissions_ before you were born.
--
73, Cecil http://www.qsl.net/w5dxp



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Cecil Moore wrote:

Jim Kelley wrote:

Cecil Moore wrote:
rearward-traveling momentum energy in those two waves is conserved by
changing direction to become part of a forward-traveling wave.

Yeeesh. You had it on, dog, up until that. And don't try to tell me
(again) that I'm lying that you said it.

(Remember when you wrote this? "If reflected energy makes its way back
into the final amp, it was never generated in the first place, by
definition." Hint: apply the same idea to your "rearward-traveling
momentum" and you'll have it.)

Egads Jim, exactly how much of reality do you think I am capable of
ignoring?

You're evidently capable of ignoring at least some, Cecil.

I *don't* agree with that definition above and your implication that
momentum and energy don't need to be conserved is simply metaphysics
in action.

We both know that momentum and energy must be conserved. We just
disagree agree on how nature chooses to do that. And, because of that
disagreement, I'm forced to endure your beligerant rhetoric.

73, Jim AC6XG